//不同的⼦序列（hard）: https://leetcode.cn/problems/distinct-subsequences/
class Solution {
public:
    int numDistinct(string s, string t) {
        // 1. 创建 dp 表
        // 2. 初始化
        // 3. 填表
        // 4. 返回值
        int m = t.size(), n = s.size();
        vector<vector<double>> dp(m + 1, vector<double>(n + 1));
        for (int j = 0; j <= n; j++)
            dp[0][j] = 1; // 初始化
        for (int i = 1; i <= m; i++)
            for (int j = 1; j <= n; j++) {
                dp[i][j] += dp[i][j - 1];
                if (t[i - 1] == s[j - 1])
                    dp[i][j] += dp[i - 1][j - 1];
            }
        return dp[m][n];
    }
};

//通配符匹配（hard）: https://leetcode.cn/problems/wildcard-matching/
class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.size(), n = p.size();
        s = " " + s, p = " " + p;
        vector<vector<bool>> dp(m + 1, vector<bool>(n + 1)); // 1. 创建 dp 表
        dp[0][0] = true;                                     // 2. 初始化
        for (int j = 1; j <= n; j++)

        if (p[j] == '*')
            dp[0][j] = true;
        else
            break;
        // 3. 填表
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (p[j] == '*')
                    dp[i][j] = dp[i - 1][j] || dp[i][j - 1];
                else
                    dp[i][j] =
                        (p[j] == '?' || s[i] == p[j]) && dp[i - 1][j - 1];
            }
        }
        // 4. 返回值
        return dp[m][n];
    }
};